I tried it before watching. If you want to solve it on hard mode, introduce α in the ln with 2+αx only in the numerator (so I(-1) = 0) then try to solve it. I went through two separate Feynman techniques then did an integration by parts to get dialogarithms before I found an error in the beginning of my work, where i forgot to divide by two. I don't want to re-do most of the work, but it seems possible that way.
Euler substitution sqrt(1-x^2) = 1 - xt We will have Int(ln((t^2+t+1)/(t^2-t+1))/t,t=0..1) Integration by parts with u = ln((t^2+t+1)/(t^2-t+1)) dv = 1/t dt We have integral 2Int(ln(t)*(t^2-1)/(t^4+t^2+1),t=0..1) Expand rational factor 2Int(ln(t)*(t^2-1)^2/((t^4+t^2+1)(t^2-1)),t=0..1) 2Int((t^4-2t^2+1)*ln(t)/(t^6-1),t=0..1) =-2Int((t^4-2t^2+1)*ln(t)/(1-t^6),t=0..1) Now we expand 1/(1-t^6) as geometric series with common ratio t^6 -2(Int(sum(t^(6n+4)*ln(t),n=0..infinity) -sum(t^(6n+2)*ln(t),n=0..infinity) + sum(t^(6n)*ln(t),n=0..infinity) ,t=0..1) Here we change order of integration and summation then we calculate integrals by parts Finally we will have three sums to evaluate -2(sum(-1/(6n+1)^2,n=0..infinity)-2*sum(-1/(6n+3)^2,n=0..infinity)+sum(-1/(6n+5)^2,n=0..infinity)) -2(sum(-1/(6n+1)^2,n=0..infinity)+sum(-1/(6n+3)^2,n=0..infinity)+sum(-1/(6n+5)^2,n=0..infinity) - 3sum(-1/(6n+3)^2,n=0..infinity)) -2(sum(-1/(2n+1)^2,n=0..infinity)-1/3sum(-1/(2n+1)^2,n=0..infinity)) 2(sum(1/(2n+1)^2,n=0..infinity)-1/3sum(1/(2n+1)^2,n=0..infinity)) 4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*(sum(1/(n+1)^2,n=0..infinity) - sum(1/(4(n+1)^2),n=0..infinity)) 4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*(sum(1/(n+1)^2,n=0..infinity) - 1/4*sum(1/(n+1)^2,n=0..infinity)) 4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*3/4*sum(1/(n+1)^2,n=0..infinity) 4/3*sum(1/(2n+1)^2,n=0..infinity)=sum(1/n^2,n=1..infinity) =π^2/6
use the fact that integral from 0 to 1 of f(x) = integral from 0 to 1 of f(1-x) to get 1-x in denominator - then use the series expansion of 1/1-x and maybe we get something
y=ln((1+x)/(1-x)) e^y=(1+x)/(1-x) e^y-xe^y=1+x (1+e^y)x+(1-e^y)=0 x=(e^y-1)/(e^y+1) x=tanh(y/2) y=2artanh(x) ln((1+x)/(1-x))=2artanh(x) not sure what you can do with this but it's something i noticed
9:20 my mind blown bro after seeing the trick of looking up anti derivatives which is not happened quite a while pardon me not quite almost never (except some differential equation videos)😅 😊😂😂
Btw do i always use lim as parameter goes to infinity for the part of the function where i inserted parameter or i can use whatever number for example limit as parameter goes to 0,1,-2... edit: when i want to find the missing constant c
You should use any "convenient" number for that purpose. In this case it happened to be infinity, which isn't a number you can just substitute in so you have to consider it as a limit.
@@sadi_supercell2132 if you're left with an expression like that it means that you didn't choose a good parameterization to begin with. You should always choose where to insert the "a" to end up with an expression that is "nice".
Pi^2/6 makes me think theres gotta be a way using series to turn it into basel somehow and now i need to find it
Did you find it?
I tried it before watching. If you want to solve it on hard mode, introduce α in the ln with 2+αx only in the numerator (so I(-1) = 0) then try to solve it. I went through two separate Feynman techniques then did an integration by parts to get dialogarithms before I found an error in the beginning of my work, where i forgot to divide by two. I don't want to re-do most of the work, but it seems possible that way.
Wow now that's some roughhousing😂
Bro solved a monster integral and when he had to evaluate 1/2-1/3 he said this is the hard part 😭💀
If you rewrite ln[(2+x)/(2-x)] as a series combine and switch them sums, we can rewrite the integral as the beta function.
1:10 when alpha=inf it approch zero
Hence you can straight away intergrate the whole thing from [2,inf)11:43
the "constant of integration C" line made me piss my trousers
😂😂😂
Yes, indeed, write ln[(2+x)/(2-x)] = ln(1+ x/2) - ln(1-x/2) = 2*argth(x/2) and expand those ln's in taylor series as for 0
*uses the overpowered technique of looking up a table of antiderivatives*
Very nice solution. Thank you
12：42 Fourrier series:
1/1²+1/2²+1/3²+1/4²+....=pi²/6
8:58 He got me here
can you do a follow up video on this very modern technique at 9:20?
looking up the table on anti derivatives? its basically a result you have to keep in mind like how integral of arcsin x is 1/sqrt(1-x^2)
@@blibilbbro he is joking
don't u get it
that he understands all the other awesome tricks but not this single step😂😂😂
Euler substitution
sqrt(1-x^2) = 1 - xt
We will have
Int(ln((t^2+t+1)/(t^2-t+1))/t,t=0..1)
Integration by parts with
u = ln((t^2+t+1)/(t^2-t+1)) dv = 1/t dt
We have integral
2Int(ln(t)*(t^2-1)/(t^4+t^2+1),t=0..1)
Expand rational factor
2Int(ln(t)*(t^2-1)^2/((t^4+t^2+1)(t^2-1)),t=0..1)
2Int((t^4-2t^2+1)*ln(t)/(t^6-1),t=0..1)
=-2Int((t^4-2t^2+1)*ln(t)/(1-t^6),t=0..1)
Now we expand 1/(1-t^6) as geometric series with common ratio t^6
-2(Int(sum(t^(6n+4)*ln(t),n=0..infinity) -sum(t^(6n+2)*ln(t),n=0..infinity) + sum(t^(6n)*ln(t),n=0..infinity) ,t=0..1)
Here we change order of integration and summation
then we calculate integrals by parts
Finally we will have three sums to evaluate
-2(sum(-1/(6n+1)^2,n=0..infinity)-2*sum(-1/(6n+3)^2,n=0..infinity)+sum(-1/(6n+5)^2,n=0..infinity))
-2(sum(-1/(6n+1)^2,n=0..infinity)+sum(-1/(6n+3)^2,n=0..infinity)+sum(-1/(6n+5)^2,n=0..infinity) - 3sum(-1/(6n+3)^2,n=0..infinity))
-2(sum(-1/(2n+1)^2,n=0..infinity)-1/3sum(-1/(2n+1)^2,n=0..infinity))
2(sum(1/(2n+1)^2,n=0..infinity)-1/3sum(1/(2n+1)^2,n=0..infinity))
4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*(sum(1/(n+1)^2,n=0..infinity) - sum(1/(4(n+1)^2),n=0..infinity))
4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*(sum(1/(n+1)^2,n=0..infinity) - 1/4*sum(1/(n+1)^2,n=0..infinity))
4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*3/4*sum(1/(n+1)^2,n=0..infinity)
4/3*sum(1/(2n+1)^2,n=0..infinity)=sum(1/n^2,n=1..infinity)
=π^2/6
use the fact that integral from 0 to 1 of f(x) = integral from 0 to 1 of f(1-x) to get 1-x in denominator - then use the series expansion of 1/1-x and maybe we get something
1operazione..x=sinθ...{...ln (2+sinθ/2-sinθ)*1/sinθ..}….uso feyman con I(a)={...2-asinθ/2+asinθ...}..I(0)=0,I=I(1)...risulta I'(a)=π/√(4-a^2),I=πarcsin(a/2)...I=I(1)=πarcsin(1/2)=π^2/6
isnt there a mistake at 5:30? why do we just assume we can distribute a sec^2 over (cos^2 +a^2-1)?
Bro there is a u beside the left parentheses
@@venkatamarutiramtarigoppul2078 oh my bad i just misread it
so does I(z) = -pi*sec-1(z) + pi^2/2 = Z(z) for any complex z?
y=ln((1+x)/(1-x))
e^y=(1+x)/(1-x)
e^y-xe^y=1+x
(1+e^y)x+(1-e^y)=0
x=(e^y-1)/(e^y+1)
x=tanh(y/2)
y=2artanh(x)
ln((1+x)/(1-x))=2artanh(x)
not sure what you can do with this but it's something i noticed
isnt x = e^y-1/e^y+1 ?
@@anupamamehra6068 oh true sign error mb
@@anupamamehra6068 fixed
I think I missed the part of the solution where you got the cosecant and cotangent functions in as well.
9:20 my mind blown bro after seeing the trick of looking up anti derivatives
which is not happened quite a while
pardon me not quite almost never (except some differential equation videos)😅
😊😂😂
Yes, ζ(s) = π^2 /2 - π arcsec(s)
I went through all your integration problems but first time I understood it
Additional observation: when we take the integration limits from 1 to 2 we get -5/3*i*Gieseking's constant.
Real beauty!
Bro can you try proving this one? integral 0 to pi arctan(ln(sinx)/x) dx = - pi*arctan[2ln(2)/pi]
Btw do i always use lim as parameter goes to infinity for the part of the function where i inserted parameter or i can use whatever number for example limit as parameter goes to 0,1,-2... edit: when i want to find the missing constant c
You should use any "convenient" number for that purpose. In this case it happened to be infinity, which isn't a number you can just substitute in so you have to consider it as a limit.
@@zunaidparkerbut here is the problem , which number i substitute in ln(a(x^2+1)) so i get the 0 for the limit
@@zunaidparker maybe a=1 for ln(a) + ln(x^2+1) but what do i do for ln(x^2+1)
@@sadi_supercell2132 if you're left with an expression like that it means that you didn't choose a good parameterization to begin with. You should always choose where to insert the "a" to end up with an expression that is "nice".
@@zunaidparker integral from 0 to infinity of ln(x^2+1)/x^2+1
asnwer=1+2 /1-/2
Ohh My god!😮
Beautifull
so coooool！
❤❤❤
Bro why do some of your video titles read like they're from the hub.